UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
PHY3703
Statistical and Thermal Physics
Formative Assessment 3
Prescribed Textbook: Harvey Gould and Jan Tobochnik,
Statistical and Thermal Physics with Computer Applications,
Princeton University Press, 2010
Problems Covered: 5.16, 5.26, 6.1, 6.3
Submission Date: As per myUnisa
Submitted in partial fulfilment of the requirements for PHY3703
, Statistical and Thermal Physics PHY3703 – Assessment 3
Problem 5.16 Behaviour of the Specific Heat Near Tc
Question (i): Show that C(T → Tc− ) = 3k/2
Answer:
From mean-field theory, Eq. (5.120) gives the internal energy per spin as a function of
magnetisation m. The relevant term near Tc is proportional to m2 , so the energy per
particle takes the form:
1
U = U0 − Jz m2
2
where J is the exchange coupling and z is the coordination number. Near the critical
point, the given approximation states:
3(Tc − T )
m2 ≈ , T ≲ Tc
Tc
Substituting:
1 3(Tc − T ) 3Jz
U ≈ U0 − Jz · = U0 − (Tc − T )
2 Tc 2Tc
Differentiating with respect to T to find the heat capacity per spin:
dU 3Jz
C= =
dT 2Tc
In mean-field theory the critical temperature satisfies kTc = Jz, so Jz = kTc . Substitut-
ing:
3kTc
C=
2Tc
3
C(T → Tc− ) = k
2
Question (ii): Show that mean-field theory predicts a jump (discontinuity) in
C
Answer:
The specific heat behaves differently on either side of Tc .
Above Tc (T > Tc ): The system is in the disordered (paramagnetic) phase with m = 0.
There is no contribution to the energy from magnetic ordering, and from Eq. (5.120) the
heat capacity per spin is:
1
College of Science, Engineering and Technology
PHY3703
Statistical and Thermal Physics
Formative Assessment 3
Prescribed Textbook: Harvey Gould and Jan Tobochnik,
Statistical and Thermal Physics with Computer Applications,
Princeton University Press, 2010
Problems Covered: 5.16, 5.26, 6.1, 6.3
Submission Date: As per myUnisa
Submitted in partial fulfilment of the requirements for PHY3703
, Statistical and Thermal Physics PHY3703 – Assessment 3
Problem 5.16 Behaviour of the Specific Heat Near Tc
Question (i): Show that C(T → Tc− ) = 3k/2
Answer:
From mean-field theory, Eq. (5.120) gives the internal energy per spin as a function of
magnetisation m. The relevant term near Tc is proportional to m2 , so the energy per
particle takes the form:
1
U = U0 − Jz m2
2
where J is the exchange coupling and z is the coordination number. Near the critical
point, the given approximation states:
3(Tc − T )
m2 ≈ , T ≲ Tc
Tc
Substituting:
1 3(Tc − T ) 3Jz
U ≈ U0 − Jz · = U0 − (Tc − T )
2 Tc 2Tc
Differentiating with respect to T to find the heat capacity per spin:
dU 3Jz
C= =
dT 2Tc
In mean-field theory the critical temperature satisfies kTc = Jz, so Jz = kTc . Substitut-
ing:
3kTc
C=
2Tc
3
C(T → Tc− ) = k
2
Question (ii): Show that mean-field theory predicts a jump (discontinuity) in
C
Answer:
The specific heat behaves differently on either side of Tc .
Above Tc (T > Tc ): The system is in the disordered (paramagnetic) phase with m = 0.
There is no contribution to the energy from magnetic ordering, and from Eq. (5.120) the
heat capacity per spin is:
1
